integral from 0 to 2pi of cos(2x)sin(5x)

解

\int_{0}^{2 π} cos (2 x) sin (5 x) d x

0

解答ステップ

\int_{0}^{2 π} cos (2 x) sin (5 x) d x

次の恒等を使用する:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int_{0}^{2 π} \frac{sin ( 5 x + 2 x ) + sin ( 5 x - 2 x )}{2} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{2 π} sin (5 x + 2 x) + sin (5 x - 2 x) d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{2 π} sin (5 x + 2 x) d x + \int_{0}^{2 π} sin (5 x - 2 x) d x)

\int_{0}^{2 π} sin (5 x + 2 x) d x = 0

\int_{0}^{2 π} sin (5 x - 2 x) d x = 0

= \frac{1}{2} (0 + 0)

簡素化

= 0